【模板】二逼平衡树

题目链接：传送门
题解：线段树套splay

//by sdfzchy#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define ls t[x].ch[0]#define rs t[x].ch[1]#define pa t[x].fausing namespace std;typedef long long LL;const int inf=(1<<30),N=3000010,mod=1e9+7;int n,m,a[N];struct node{    int ch[2],fa,siz,cnt,val;}t[N];struct seg{    int l,r,rt;}c[N];inline int in(){    char tmp=getchar();    int res=0,f=1;    while((tmp<'0'||tmp>'9')&&tmp!='-')tmp=getchar();    if(tmp=='-') f=-1,tmp=getchar();    while(tmp>='0'&&tmp<='9')   res=(res<<1)+(res<<3)+(tmp^48),tmp=getchar();    return res*f;}inline void upd(int x) {t[x].siz=t[ls].siz+t[rs].siz+t[x].cnt;}inline int gi(int x) {return t[pa].ch[1]==x;}inline void rot(int x){    int f=pa,g=t[f].fa,o=gi(x);    if(g) t[g].ch[gi(f)]=x;t[x].fa=g;    t[f].ch[o]=t[x].ch[!o];t[t[x].ch[!o]].fa=f;    t[x].ch[!o]=f;t[f].fa=x;    upd(f),upd(x);}int cur,cnt,ans,tmp;inline void splay(int x,int tar,int &root){    for(;pa!=tar;rot(x))        if(t[pa].fa!=tar)            rot((gi(x)==gi(pa))?pa:x);    if(!tar) root=x;}inline void newnode(int &x,int val){    x=++cnt;    ls=rs=0;    t[x].siz=t[x].cnt=1;    t[x].val=val;}void ins(int &x,int fa,int w){    int k=x;    if(!x) {newnode(x,w);pa=fa;return;}    t[x].siz++;    if(t[x].val==w) {t[x].cnt++;return;}    else if(w>t[x].val) ins(rs,x,w);    else ins(ls,x,w);}void rank(int x,int k){    while(x)    {        if(k<=t[x].val) x=ls;        else tmp+=t[ls].siz+t[x].cnt,x=rs;    }}int kth(int x,int k){    if(k<t[x].val) return kth(ls,k);    else if(k==t[x].val) return x;    else return kth(rs,k);}void del(int &root,int val){    int x=kth(root,val);    splay(x,0,root);    if(t[x].cnt>1) {t[x].cnt--;t[x].siz--;return;}    if(!ls&&!rs) root=0;    else if(rs==0) t[ls].fa=0,root=ls;    else if(ls==0) t[rs].fa=0,root=rs;    else    {        int pos=ls;        while(t[pos].ch[1]) pos=t[pos].ch[1];        splay(pos,x,root);        t[pos].ch[1]=rs;t[rs].fa=pos;        t[pos].fa=0;root=pos;        upd(root);    }}void change(int pos,int rt,int val,int las){    del(c[rt].rt,las);    ins(c[rt].rt,0,val);    int l=c[rt].l,r=c[rt].r,mid=(l+r)>>1;    if(l==r) return;    if(pos<=mid) change(pos,rt<<1,val,las);    else change(pos,rt<<1|1,val,las);}void build(int l,int r,int rt){    c[rt].l=l;c[rt].r=r;    for(int i=l;i<=r;i++) ins(c[rt].rt,0,a[i]);    if(l==r) return;    int mid=(l+r)>>1;    build(l,mid,rt<<1);    build(mid+1,r,rt<<1|1);}void pre(int x,int k){    if(!x) return;    if(t[x].val<k) ans=max(ans,t[x].val),pre(rs,k);    else pre(ls,k); }void suc(int x,int k){    if(!x) return;    if(t[x].val>k) ans=min(ans,t[x].val),suc(ls,k);    else suc(rs,k);}void gik(int x,int y,int rt,int k){    int l=c[rt].l,r=c[rt].r,mid=(l+r)>>1;    if(x==l&&r==y) rank(c[rt].rt,k);    else if(x>mid) gik(x,y,rt<<1|1,k);    else if(y<=mid) gik(x,y,rt<<1,k);    else gik(x,mid,rt<<1,k),gik(mid+1,y,rt<<1|1,k);}void gipre(int x,int y,int rt,int k){    int l=c[rt].l,r=c[rt].r,mid=(l+r)>>1;    if(x==l&&r==y) pre(c[rt].rt,k);    else if(x>mid) gipre(x,y,rt<<1|1,k);    else if(y<=mid) gipre(x,y,rt<<1,k);    else gipre(x,mid,rt<<1,k),gipre(mid+1,y,rt<<1|1,k); }void gisuc(int x,int y,int rt,int k){    int l=c[rt].l,r=c[rt].r,mid=(l+r)>>1;    if(x==l&&r==y) suc(c[rt].rt,k);    else if(x>mid) gisuc(x,y,rt<<1|1,k);    else if(y<=mid) gisuc(x,y,rt<<1,k);    else gisuc(x,mid,rt<<1,k),gisuc(mid+1,y,rt<<1|1,k); }int main(){    n=in(),m=in();    for(int i=1;i<=n;i++) a[i]=in();    build(1,n,1);    for(int i=1,op,l,r,z;i<=m;i++)    {        op=in(),l=in(),r=in();        switch(op)        {            case 1:z=in();tmp=1;gik(l,r,1,z);printf("%d\n",tmp);break;            case 2:                {                   z=in();                   int L=0,R=2147483647;                   while(L<=R)                   {                       int mid=(L+R)>>1;                       tmp=1;                       gik(l,r,1,mid);                       if(tmp<=z) L=mid+1,ans=mid;                       else R=mid-1;                   }                   printf("%d\n",ans);                   break;                }            case 3:change(l,1,r,a[l]);a[l]=r;break;            case 4:z=in();ans=-2147483647;gipre(l,r,1,z);printf("%d\n",ans);break;            case 5:z=in();ans=2147483647;gisuc(l,r,1,z);printf("%d\n",ans);break;        }    }    /*  for(int i=1;i<=n;i++)        {            ans=2147483647;            gisuc(i,i,1,0);            printf("%d ",ans);          }         cout<<endl;*/    return 0;}